How to calculate qh and qc?

In thermodynamics, a heat engine is a system that performs the conversion of heat or thermal energy to mechanical work.

In thermodynamics, a heat engine is a system that performs the conversion of heat or thermal energy to mechanical work. Gasoline and diesel engines, jet engines, and steam turbines are all heat engines that do work by using part of the heat transfer from some source. Heat transfer from the hot object (or hot reservoir) is denoted as Qh, while heat transfer into the cold object (or cold reservoir) is Qc, and the work done by the engine is W. The temperatures of the hot and cold reservoirs are Th and Tc, respectively.

Heat Transfer: (a) Heat transfer occurs spontaneously from a hot object to a cold one, consistent with the second law of thermodynamics. (b) A heat engine, represented here by a circle, uses part of the heat transfer to do work. The hot and cold objects are called the hot and cold reservoirs. Qh is the heat transfer out of the hot reservoir, W is the work output, and Qc is the heat transfer into the cold reservoir.

Thermodynamics and Heat Engines: A brief introduction to heat engines and thermodynamic concepts such as the Carnot Engine for students.

In thermodynamics, a heat engine is a system that performs the conversion of heat or thermal energy to mechanical work.

In thermodynamics, a heat engine is a system that performs the conversion of heat or thermal energy to mechanical work. Gasoline and diesel engines, jet engines, and steam turbines are all heat engines that do work by using part of the heat transfer from some source. Heat transfer from the hot object (or hot reservoir) is denoted as Qh, while heat transfer into the cold object (or cold reservoir) is Qc, and the work done by the engine is W. The temperatures of the hot and cold reservoirs are Th and Tc, respectively.

Heat Transfer: (a) Heat transfer occurs spontaneously from a hot object to a cold one, consistent with the second law of thermodynamics. (b) A heat engine, represented here by a circle, uses part of the heat transfer to do work. The hot and cold objects are called the hot and cold reservoirs. Qh is the heat transfer out of the hot reservoir, W is the work output, and Qc is the heat transfer into the cold reservoir.

Device that converts thermal E. (Qh) into mechanical or electrical E.

During the cycle,

Heat (Qh) is absorbed from a source at a high T.

Work is done by the engine.

Heat (Qc) is expelled to a source at a lower T.

Since it is a cyclic process, ∆U = 0. Thus, W = Qtotal ---> W = Qh – Qc

η = efficiency = ratio of work done to heat absorbed

η = W / Qh = (Qh – Qc) / Qh = 1 – Qc / Qh

η = 100% only if Qc = 0. i.e. a perfect heat engine would convert all of the absorbed heat Qh into mechanical work. From 2nd law of thermo, this is impossible: only a fraction of heat is converted to mechanical E.

E.g. η of automobile engines = 20%.

Kelvin–Planck form of the 2nd law of thermodynamics:

“It is impossible by a cycle to take heat from a hot reservoir and convert it into work without, at the same time, transferring an amount of heat from hot to cold reservoir.”

Device that converts work into thermal E. (Qh).

Used to heat/cool homes.

During the cycle,

Heat (Qc) is absorbed from a source at low T (e.g. outside air or food) by a circulating fluid.

Work is done on the engine by a compressor.

Heat (Qh) is expelled to a source at higher T (e.g. room).

Since it is a cyclic process, ∆U = 0. Thus, W = Qtotal ---> W = Qh – Qc

K = coefficient of performance = ratio of heat transferred to work required to transfer the heat.

K = Qh / W = Qh / (Qh – Qc)

K can be much larger than 100% (denominator < 1).

Device that converts work into thermal E. (Qc).

During the cycle,

Heat (Qc) is absorbed from a source at low T (e.g. outside air or food) by a circulating fluid.

Work is done on the engine by a compressor.

Heat (Qh) is expelled to a source at higher T (e.g. room).

Since it is a cyclic process, ∆U = 0. Thus, W = Qtotal  W = Qh – Qc

K = coefficient of performance = ratio of heat transferred to work required to transfer the heat.

K = Qc / W = Qc / (Qh – Qc)

K can be much larger than 100% (denominator < 1).

A perfect refrigerator would transfer heat from a colder body to a hotter body without doing any work. From 2nd law of thermo, this is impossible.

E.g. K of a refrigerator = 5 or 6

Clausius form of the 2nd law of thermo:

“It is impossible to use a cyclic process to transfer heat from a colder to a hotter body without doing work on the system.”

i.e. heat will not flow spontaneously from a cold to a hot object.

• Carnot made the most efficient heat engine. Net work taken from the Carnot cycle is the largest possible for a given amount of heat supplied.

• Carnot’s theorem: No real (irreversible) engine operating between 2 heat reservoirs can be more efficient than a Carnot (reversible) engine operating between the same 2 reservoirs.

• Carnot used an ideal gas contained in a cylinder with a movable piston at one end. The cylinder walls and the piston are thermally non–conducting.

• The Carnot cycle consists of 2 adiabatic and 2 isothermal processes, all reversible:

A ---> B

Isothermal expansion of a gas placed in thermal contact with a heat reservoir at temp. Th.

During the process, the gas absorbs heat Qh from the base of the cylinder and does a work WAB in raising the piston.

B ---> C

Adiabatic expansion by replacing the base of the cylinder by a thermally non–conducting wall.

During the process, T falls from Th to Tc and the gas does work WBC in raising the piston.

C ---> D

Isothermal compression by placing the gas in thermal contact with a heat reservoir at temp. Tc.

During the process, the gas expels heat Qc to the reservoir and the work WCD is done on the gas by external agent.

D ---> A

Adiabatic compression by replacing the base of the cylinder by a non–conducting wall.

During the process, T increases from Tc to Th and the work WDA is done on the gas by external agent.

• Net work done in this reversible cyclic process = area enclosed by the path ABCDA = net heat transferred into the system, since ∆U = 0.

• Since the internal E. of an ideal gas depends only on absolute T, then in AB and CD, T and hence U remain constant. From the 1st law of thermo,

Q h = W A B = n   R   T h l n V 2 V 1 {\displaystyle Q_{h}=W_{AB}=n\ R\ T_{h}ln{\frac {V_{2}}{V_{1}}}}

Q c = − W C D = n   R   T c l n V 3 V 4 {\displaystyle Q_{c}=-W_{CD}=n\ R\ T_{c}ln{\frac {V_{3}}{V_{4}}}}

• By dividing the equations,

Q h Q c = T h l n ( V 2 V 1 ) T c l n ( V 3 V 4 ) {\displaystyle {\frac {Q_{h}}{Q_{c}}}={\frac {T_{h}ln({\frac {V_{2}}{V_{1}}})}{T_{c}ln({\frac {V_{3}}{V_{4}}})}}}

• For BC and DA,

T h   V 2 ( γ − 1 ) = T c   V 3 ( γ − 1 ) {\displaystyle T_{h}\ V_{2}^{(\gamma -1)}=T_{c}\ V_{3}^{(\gamma -1)}}

T h V 1 ( γ − 1 ) = T c   V 4 ( γ − 1 ) {\displaystyle T_{h}V_{1}^{(\gamma -1)}=T_{c}\ V_{4}^{(\gamma -1)}}

• By dividing the equations and taking the ( γ − 1 ) {\displaystyle (\gamma -1)} th root,

V 3 V 4 = V 2 V 1 {\displaystyle {\frac {V_{3}}{V_{4}}}={\frac {V_{2}}{V_{1}}}}

• Thus,

Q h Q c = T h T c {\displaystyle {\frac {Q_{h}}{Q_{c}}}={\frac {T_{h}}{T_{c}}}}

η = 1 − T c T h {\displaystyle \eta =1-{\frac {T_{c}}{T_{h}}}}

(And this of course applies also to heat pumps and refrigerators. i.e., in their formulas, you can switch the Qh and Qc by Th and Tc respectively)

 Efficiency of a Carnot engine is:

η = 1 – (Qc / Qh) = 1 – (Tc / Th)

 The zero point of the thermodynamic scale is fixed as the T of the cold reservoir Tc at which η = 1 and hence Qc = 0 and W = Qh.

 T must be the absolute scale because otherwise it may have –ve values, and hence the engine will perform work more than the heat given by the source. i.e. we would create E. from nothing, which is in contradiction to 1st law of thermo.

 Thus, T = 0 is the lowest T in all scales i.e. the absolute zero.

• Isolated systems and physical processes tend towards disorder, and entropy is a measure of this disorder.

• E.g. If all molecules of a gas in a room move together, this is a very ordered, unlikely state. If the molecules move randomly in all directions, changing speed after collisions, this is a very disordered, likely state.

• In a reversible process between 2 equilibrium states, change in entropy is given by:

dS = dQr / T

Where dS = change in entropy, dQr = heat absorbed or expelled by the system in reversible process, T = absolute T.

• Heat absorbed by the system = +ve dQr and S↑. Heat lost = –ve dQr and S↓.

• The most useful statement of the 2nd law of thermodynamics:

• For reversible adiabatic process or reversible reactions, ∆S = 0.

For irreversible process or irreversible reactions, ∆S > 0.

Where ∆S = change in entropy of the system + surroundings (the universe).

∆S = ∫dS = ∫dQr / T

For reversible adiabatic process, no heat is transferred between system and surroundings, so ∆S = 0.

For Carnot engine, ∆S = Qh/Th – Qc/Tc. Since Qc/Qh = Tc/Th, then ∆S = 0.

For an ideal gas undergoing a quasi–static reversible process from Ti Vi to Tf Vf,

dQr = dU + dW  ; dW = PdV. Since it’s n ideal gas,

dU = n cv dT  ; P = n R T / V. Thus,

dQr = n cv dT + n R T dV / V. Dividing by T,

dQr / T = n cv (dT / T) + n R (dV / V). Assuming cv is constant,

by integration, ***

∆S = ∫dQr / T = n cv ln (Tf / Ti) + n R ln (Vf / Vi)

Thus, ∆S is independent of the reversible path and depends only on initial and final states

For cyclic process, Ti = Tf and Vi = Vf, so ∆S = 0.

 When heat transfers from a hot Th to cold Tc reservoirs, entropy of cold reservoir increases by Q/Tc and entropy of hot reservoir decreases by Q/Th.

 Since Tc < Th, total change in entropy of the system (universe) > 0

∆Su = Q / Tc – Q / Th > 0

 Thus, if ∆Su < 0, process cannot occur (e.g. heat transfer from cold to hot body).

• An ideal gas in an insulated container occupies a volume Vi. A membrane separating it from a vacuum is suddenly broken and the gas expands irreversibly to Vf.

• Work done against the vacuum = 0. Since walls are insulating, Q = 0. Thus, ∆U = 0 and Ui = Uf. Since the gas is ideal, U depends only on T. Thus, Ti = Tf.

• Since ∆S = ∫dS = ∫dQr / T only applies to reversible processes, we cannot use it directly. Thus, we imagine a reversible process with the same initial and final states: an isothermal reversible expansion. Since T is constant, ∆S = ∫dQr / T = 1/T ∫dQr.

• Since it is isothermal, ∫dQr = W = n R T ln (Vf / Vi). Thus,

∆S = n R ln (Vf / Vi)

Since Vf > Vi, ∆Su > 0. This can also be obtained be eqn *** by putting Tf = Ti